Introduction
This tutorial examines the flow of liquid metal through a two dimensional channel. The viscosity of the liquid metal is modeled as a function of the temperature using a user-defined function.
Problem Description
The problem considered in this tutorial is shown schematically in Figure. As the symmetry condition is imposed at the center line, only half the channel is modeled, The wall of the channel is split into two parts: wall-2, which has a temperature of 280 K and wall-3 which has a temperature of 290 K. The temperature-dependent viscosity of the liquid metal will respond to this change in wall temperature. The function, named cell viscosity, is defined on a cell using DEFINE PROPERTY. Two real variables are introduced: temp, the value of C_T(cell, thread), and mu, the laminar viscosity computed by the function. The value of the temperature is checked, and based upon the range into which it falls, the appropriate value of mu is computed. At the end of the function, the computed value for mu is returned to the solver.
This tutorial examines the flow of liquid metal through a two dimensional channel. The viscosity of the liquid metal is modeled as a function of the temperature using a user-defined function.
Problem Description
The problem considered in this tutorial is shown schematically in Figure. As the symmetry condition is imposed at the center line, only half the channel is modeled, The wall of the channel is split into two parts: wall-2, which has a temperature of 280 K and wall-3 which has a temperature of 290 K. The temperature-dependent viscosity of the liquid metal will respond to this change in wall temperature. The function, named cell viscosity, is defined on a cell using DEFINE PROPERTY. Two real variables are introduced: temp, the value of C_T(cell, thread), and mu, the laminar viscosity computed by the function. The value of the temperature is checked, and based upon the range into which it falls, the appropriate value of mu is computed. At the end of the function, the computed value for mu is returned to the solver.
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